Two identical blocks are placed on a tabletop on either side of a spring, and sq

Question

Two identical blocks are placed on a tabletop on either side of a spring, and squeezed together to compress the spring to a length of L. The blocks are then suddenly released so that the spring expands and the blocks shoot apart. The blocks each have mass m, and the spring has stiffness k and relaxed length L0.

(a) If the tabletop is frictionless, how fast are the blocks moving when they leave the spring?

(b) If the tabletop is not frictionless, and the coefficients of static and kinetic friction between the blocks and tabletop are s and k respectively, how far apart are the blocks when they come to rest?

Explanation / Answer

The elastic potential energy stored inside the spring during the compression

Elastic potential energy Ep = ½ kx2

Ep = ½ k (L-Lo)2

This energy is then transferred to the two blocks equally and as their kinetic energy

Ek = 2 (½ mv2)

Ek = mv2

Ep = Ek

½ k (L-Lo) 2= mv2

v = (L-Lo) x (k/2m)

B)

Applied force by the spring on both masses

2F = kx = k (L-Lo)

Applied force on one mass is

F= k (L-Lo)/2

The normal force acting on one block N = mg

The static friction Fs = µsmg

The kinetic friction Fk = µkmg

The block will be at rest if Fs > F

If F >Fs

Then the lock will move with the velocity v = (L-Lo) x (k/2m)

Until all its kinetic energy is spent in doing work against friction

Work done against friction

W = Fk x D

½ mv2 = µkmgD

D = v2/2µkg

D = k (L-Lo)2/4m µkg

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